Oracle根据连续性日期的重复数据取最大或最小值日期

标签：ast   with   for   group   dual   复数   rac   form   table   

原始数据：

 结果数据：

 对比两个图，要是不处理连续性中的重复值，我们直接可以用LEAD函数了事，但处理出来的结果貌似多余。

我的思路是先将原始数据中连续性日期有重复值的处理好，即选择最小的一个，比如2011/01/01和2012/01/01的值是一样，日期又是连续的，所以要去除2012/01/01，保留2011/01/01,让处理出来的截止日期为第4行的日期-1即2012/12/30.

以下是实现代码，有更好更方便的欢迎提出来，共同学习。

/*CREATE TABLE TB(PB_DATE DATE,SAVE_RATE VARCHAR2(10),LEN_RATE VARCHAR2(10));
INSERT INTO TB SELECT DATE‘2010-01-01‘,‘8%‘,‘10%‘ FROM DUAL;
INSERT INTO TB SELECT DATE‘2011-01-01‘,‘5%‘,‘7%‘ FROM DUAL;
INSERT INTO TB SELECT DATE‘2012-01-01‘,‘5%‘,‘7%‘ FROM DUAL;
INSERT INTO TB SELECT DATE‘2012-10-31‘,‘5%‘,‘6%‘ FROM DUAL;
INSERT INTO TB SELECT DATE‘2013-01-01‘,‘5%‘,‘6%‘ FROM DUAL;
INSERT INTO TB SELECT DATE‘2013-03-31‘,‘8%‘,‘9%‘ FROM DUAL;
INSERT INTO TB SELECT DATE‘2013-09-01‘,‘8%‘,‘10%‘ FROM DUAL;
INSERT INTO TB SELECT DATE‘2014-01-01‘,‘8%‘,‘9%‘ FROM DUAL;
INSERT INTO TB SELECT DATE‘2015-01-01‘,‘6%‘,‘9%‘ FROM DUAL;
SELECT MIN(PB_DATE)START_DATE,MAX(END_DATE)END_DATE,SAVE_RATE,LEN_RATE FROM(
SELECT PB_DATE,LEAD(PB_DATE,1,DATE‘9999-12-31‘)OVER(ORDER BY PB_DATE)-1 AS END_DATE,SAVE_RATE,LEN_RATE FROM TB)
GROUP BY SAVE_RATE,LEN_RATE;
SELECT T.*,MAX(PB_DATE)KEEP(DENSE_RANK LAST ORDER BY SAVE_RATE,LEN_RATE)OVER(PARTITION BY SAVE_RATE,LEN_RATE)AS RN FROM TB T ORDER BY 1 ;
SELECT * FROM TB;*/
/* Formatted on 2019/9/24 12:29:14 (QP5 v5.227.12220.39754) */
/* Formatted on 2019/9/24 12:41:16 (QP5 v5.227.12220.39754) */

实现代码，以上代码是造数脚本：
WITH TMP
AS ( SELECT PB_DATE,
SAVE_RATE,
LEN_RATE,
LEAD (PB_DATE, 1, DATE ‘9999-12-31‘) OVER (ORDER BY PB_DATE)
NEXT_DATE,
LAG (PB_DATE, 1, PB_DATE) OVER (ORDER BY PB_DATE) PRE_DATE,
LEAD (SAVE_RATE) OVER (ORDER BY PB_DATE) NEXT_SAVE,
LAG (SAVE_RATE) OVER (ORDER BY PB_DATE) PRE_SAVE,
LEAD (LEN_RATE) OVER (ORDER BY PB_DATE) NEXT_LEN,
LAG (LEN_RATE) OVER (ORDER BY PB_DATE) PRE_LEN
FROM TB
ORDER BY PB_DATE, SAVE_RATE, LEN_RATE),
TMP2
AS ( SELECT MAX (T.PB_DATE) AS PB_DATE, T.SAVE_RATE, T.LEN_RATE
FROM TMP T
WHERE (CASE
WHEN (SAVE_RATE = NEXT_SAVE AND LEN_RATE = NEXT_LEN)
OR (SAVE_RATE = PRE_SAVE AND LEN_RATE = PRE_LEN)
THEN
1
ELSE
0
END) = 1
GROUP BY T.SAVE_RATE, T.LEN_RATE
ORDER BY 1),
TMP3
AS (SELECT PB_DATE，SAVE_RATE,LEN_RATE
FROM TB T
WHERE NOT EXISTS
(SELECT NULL
FROM TMP2 R
WHERE R.SAVE_RATE = T.SAVE_RATE
AND R.LEN_RATE = T.LEN_RATE
AND R.PB_DATE = T.PB_DATE))
SELECT PB_DATE AS START_DATE,
LEAD (PB_DATE, 1, DATE ‘9999-12-31‘) OVER (ORDER BY PB_DATE) - 1
AS END_DATE,
SAVE_RATE,
LEN_RATE
FROM TMP3;

Oracle根据连续性日期的重复数据取最大或最小值日期

标签：ast   with   for   group   dual   复数   rac   form   table   

原文地址：https://www.cnblogs.com/guipeng/p/11577728.html